Proposition 2.2.5 ([Theorem 4.6, Proposition 4.11, Bau14]).label Let
be a diffusion operator, then:
- (1)
$L$ admits a self-adjoint extension.
- (2)
If $L$ is elliptic with smooth coefficients and there exists $\seq{h_n}\in C_{c}(\real^{d}; [0, 1])$ such that:
- (a)
$h_{n} \upto 1$ pointwise on $\real^{d}$ as $n \to \infty$.
- (b)
$||\dpn{Dh_n, ADh_n}{\real^{d}}||_{L^\infty(\real^d)}\to 0$ as $n \to \infty$.
then $L$ is essentially self-adjoint.
Proof. (1): Since $A$ is semipositive, $-L$ is a semipositive operator. Therefore $L$ admits a Friedrichs extension by Theorem 2.1.5.
(2): By Lemma 2.1.7, it is sufficient to show that there exists $\lambda > 0$ such that $(\lambda - L^{*})$ is injective.
In fact, $(\lambda - L^{*})$ is injective for all $\lambda > 0$. Fix $\lambda > 0$ and $u \in L^{2}(\real^{d})$ such that $L^{*}u = \lambda u$. In this case, since $\mu$ is equivalent to the Lebesgue measure, $Lu = \lambda u$ as distributions. By the Elliptic Regularity Theorem, $u \in C^{\infty}(\real^{d}) \cap L^{2}(\real^{d})$.
For any $h \in C_{c}^{\infty}(\real^{d}; \real)$,
Since
By the Schwarz inequality,
so
where
Therefore
Substituting $\seq{h_n}$ for $h$ and sending $n \to \infty$ yields that $\norm{Df}_{L^2(\real^d; \real^d)}= 0$. Therefore $f = 0$.$\square$